Let $X$ be the comb space given by $$ \begin{align} X ={} &\bigl\{ (x,y) \in \mathbb{R}^2 \mid 0 \leq y \leq 1 \text{ and } (x=0 \text{ or } x=\tfrac{1}{n} \text{ for some} \in \mathbb{N} \bigr\} \\ &\cup \bigl\{ (x,y) \in \mathbb{R}^2 \mid y=0 \text{ and } 0 \leq x \leq 1 \bigr\}. \end{align} $$
Let the set $A$ be given by $A = \{(0,1)\}$. Prove that $A$ is a deformation retract of $X$.
My attempt: Define the map $F:X \times [0,1] \to X$ given by $$ F(x,y,t) = \bigl( x, (1-t)y \bigr) $$ is continuous and satisfies $F(x,y,0) = (x,y),\, F(x,y,1)=(x,0)$.
After that, I am unable to proceed.
Your homotopy $F$ is a good start. It deforms $X$ to $X' = [0,1] \times \{0\}$ and keeps $X'$ fixed. Next deform $X'$ to $X'' = \{(0,0)\}$. This is possible with a homotopy keeping $X''$ fixed. With a final homotopy move $(0,0)$ to $(0,1)$. Composing these three homotopies yields a homotopy from $id_X$ to the constant map $c(x) = (0,1)$.
Note that $A$ is not a strong deformation retract of $X$.
$A$ being a deformation retract of $X$ means that $id_X$ is homotopic to a retraction of $X$ to $A$; one does not require that there exists a homotopy keeping $A$ fixed.
As we have seen, $X''$ is a strong deformation retract of $X$. This means in particular that $X$ is contractible to the point $(0,0)$. It is well-known that a contractible space is contractible to each of its points (the method to prove this is the same in the third step above).